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Question Number 143904 by nadovic last updated on 19/Jun/21

Answered by Dwaipayan Shikari last updated on 19/Jun/21

$$\frac{1}{\sqrt{\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=1e^{\pi i}=-1$$$$p=2$$$$\frac{d}{dx}{\mid }_{x=\frac{1}{4}}({xsin}^{-1}px+\frac{\sqrt{1-p^2{x}^2}}{p})$$$$=\frac{xp}{\sqrt{1-p^2{x}^2}}+{sin}^{-1}px-\frac{xp}{\sqrt{1-p^2{x}^2}}={sin}^{-1}px$$$$x=\frac{1}{4}itis{sin}^{-1}\frac{1}{2}=\frac{\pi }{6}$$

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